Three methods are currently used for cooling beverages. The first one is a "cold plate" method in which crystal ice is utilized. The "plate" is placed at the bottom of a compartment, separate from the beverage, which contains ice. The beverage runs through tubing in the ice-cooled cold plate on its way to the dispensing valve.
The second method is the "dry" method where the refrigeration evaporation coils surround the beverage tubing leading to the dispenser head.
The third method is called an "ice bath" or "ice bank" utilizes a water bath to cool the beverage tubing. The water, in turn, is cooled by the mechanical refrigeration. Ice is formed on the refrigeration coils during periods when little or no dispensing of a beverage is taking place. This ice is "banked" against peak periods when it releases its cooling energy by dissolving in water bath as it is needed to cool beverages being dispenses.
The "cold plate" cooling is the least expensive initially. However, the cost of ice, due to very low heat transfer efficiency, makes a cooling unit implementing such a method more expensive in the long run than a model with a mechanical refrigeration unit.
While the equipment cost is relatively low in the "dry" method, the unacceptably low heat transfer coefficient between refrigerant and beverage makes the dispenser that implements this method too large.
A conventional beverage cooling and supply system includes a water supply line extending in a cooling coil through an ice-water bath to a dispensing unit. A refrigeration unit causes the ice to be formed on the outside surfaces of coils. A pump forces the cooled water to circulate from the ice-water bath to the dispensing unit and back. So, different kinds of beverages are cooled by cooling means in an ice-water container.
This method is very inefficient because of the low value of heat transfer coefficient (not more than 400 W/(m.sup.2.K)) from a boiled refrigerant to a beverage. This coefficient varies in time and along cooling surface because a direct expansion (DX) refrigerant feed method is used in a refrigerant system. As the refrigerant passes through the evaporator, liquid continues to evaporate to perform cooling until only saturated vapor remains. About 75% to 85% of the heat transfer surface has been used at this point. The last 15% to 25% of the heat transfer surface is dedicated to thermal expansion valve (TXV) function and compressor protection by generating 4.4.degree. C. to 5.6.degree. C. superheat (temperature drop) in the outlet vapor. That's why in comparison to other feed techniques the DX method requires the largest heat transfer surface (by 15.div.25%), constant attention to TXV operation and great care in suction piping design to ensure compressor protection and to provide adequate oil return.
Further, film coefficients between refrigerant copper coil and water, and water and beverage stainless steel tubing are small. For example, film coefficient by water side is about 100.div.500 W/(m.sup.2.K). It is explained by the fact that water velocity along tubes is small (.apprxeq.0.2 m/s). Thus, a heat transfer between tubes and a coolant (water) by the convection process is low.
Moreover, there is large energy consumption by water ice forming on the outside surface of the coil evaporator because an every new frozen ice layer creates additional heat impedance for the next ice layer. It makes it necessary to decrease the refrigerant evaporating temperature from (-4.div.-6).degree.C. to (-12.div.-16).degree.C. At the same time every 1.degree. C. reduction of the evaporating temperature corresponds to 4% cold capacity decreasing. It means that the 10.degree. C. dropping of evaporating temperature leads to 40% decreasing of compressor cold capacity and comfortably to the 20.div.30% increasing of energy consumption by the compressor.
A further serious disadvantage of the above system is caused by long time required to prepare the cooling refrigeration unit of the dispenser. It is explained by the ineffective cooling method of a coolant (water) in a water bath volume of the dispenser. For example, for a 16 gallon water bath tank capacity, compressor power equal to 0.56 kW, and initial coolant (water) temperature 30.degree. C., the minimum duration to produce 55,6 lbs (25 kg) ice is 3.6 hours. By this ice mass, it is possible to cool the beverage peak flow 10.6 GPH (40 LPH) from 30.degree. C. to 4.degree. C. during only 1.6 hours. The summary time of the cycle is 5.2 hours. Thus, it is assumed to use the ice bank type dispenser with beverage peak flow at twenty-four hours during only a very short period equal to 1.6.times.(24:5.2).apprxeq.7.4 hour. It should be noted here that it is very important to keep the beverage final temperature in the range (+4.div.+6).degree.C. due to the following requirements:
decelerate the microorganisms growth, PA1 keep the taste of the beverage, PA1 decrease sharply a foam quantity in the time of dishes filling. PA1 N=energy, absorbed by a cooled beverage during 24 hours at peak beverage flow, known initial and final beverage temperatures, kW.h; PA1 c=mass thermal capacity of beverage, kJ/(kg..degree.C.), (f.e., for apple juice c=3.8 kJ/(kg..degree.C.)); PA1 t.sub.m -t.sub.f =temperature difference of initial and final beverage temperatures, .degree.C.; PA1 .rho.=beverage density, kg/L, (f.e., for apple juice .rho.=1.06 kg/L); PA1 f=peak beverage flow, L/H; PA1 .DELTA..tau.=work mode during 24 hours when dispenser works at peak flow, H; PA1 P--electric energy, consumed by dispenser during 24 hours, kW.H; PA1 m=ice mass produced during one cycle, kg; PA1 z=quantity of ice forming cycles during twenty-four hours; PA1 r=latent heat, 80 kcal/kg=0.093 (kW.H)/kg.
Besides, by any types of known dispensers, it is impossible to reach a temperature closed by a beverage crystallization temperature .apprxeq.(2.div.0).degree.C. without troubles or freezing of soda (mixture of drinking water and compressed vapours of dioxide of carbon).
In general, the efficiency of cooling by any type of dispenser, including ice bank type dispenser, is characterized by the following coefficient of performance (COP) EQU COP=N/P=[0.28.times.10.sup.-3 .times.f.times..rho..times.c.times.(t.sub.m -t.sub.f).times..DELTA..tau./(m.times.z.times.r)
where
According to the above mentioned data of the ice bank type dispenser action, and using the suggested formula, one can easy calculate the coefficient of performance COP.apprxeq.0.56. It appears that the ice bank type dispenser works very unsatisfactorily and expends energy more than twice as much compared to what it returns for cooling effect.